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2 years ago

Please help me asap for brainlist its for final exams

Mathematics

1 answer:

kumpel [21]2 years ago

6 0

Answer: y = — $\frac{1}{2}$ x + 3

Step-by-step explanation:

The line crosses the y-axis at the point (0,3), making its y-intercept be +3.

The line is going downwards left to right, giving it a negative slope.

The line's slope is — $\frac{1}{2}$ because it is negative and it goes down 1, over 2.

Therefore, the line's equation is y = — $\frac{1}{2}$ x + 3

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Give the domain and range of each relation X Y -3 4 -1 2 0 0 1 2 3 -4

Helga [31]

Answer:

Domain: -3, -1, 0, 1, 3

Range: 4, 2, 0, 2, -4

Step-by-step explanation:

Domain: -3, -1, 0, 1, 3

Range: 4, 2, 0, 2, -4

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3 years ago

Suppose a population contains 40000 people. All else being equal, a study based on a population sample that includes which of th

dusya [7]

Answer:

4000

Step-by-step explanation:

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3 years ago

What is the value of x? 12 units 10 15 units 20 units 24 units

vesna_86 [32]

Answer:

x = 20 units

Step-by-step explanation:

Using the Altitude- on- Hypotenuse theorem

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5 0

3 years ago

Find the constant difference for a hyperbola with foci (-3.5, 0) and (3.5, 0) and a point on the hyperbola (3. 5, 24).

Solnce55 [7]

Answer:

2a=1

Step-by-step explanation:

The constant difference for a hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is $2a.$

1. Since hyperbola has foci (-3.5, 0) and (3.5, 0), then $c=3.5.$ Note that $c=\sqrt{a^2+b^2},$ then

$\sqrt{a^2+b^2}=3.5\Rightarrow a^2+b^2=3.5^2.$

2. Since point (3.5,24) lies on the hyperbola, then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1.$

3. Solve the system of two equations:

$\left\{\begin{array}{l}a^2+b^2=3.5^2\\\dfrac{3.5^2}{a^2}-\dfrac{24^2}{b^2}=1\end{array}\right.$

From the 1st equation,

$b^2=3.5^2-a^2,$

then

$\dfrac{3.5^2}{a^2}-\dfrac{24^2}{3.5^2-a^2}=1,\\ \\\dfrac{3.5^2(3.5^2-a^2)-24^2a^2}{a^2(3.5^2-a^2)}=1,\\ \\3.5^4-3.5^2a^2-24^2a^2=3.5^2a^2-a^4,\\ \\a^4-a^2(2\cdot 3.5^2+24^2)+3.5^4=0,\\ \\a^4-600.5a^2+150.0625=0,\\ \\D=(-600.5)^2-4\cdot 150.0625=360000,\\ \\a^2_{1,2}=\dfrac{600.5\pm 600}{2}=\dfrac{1}{4},600\dfrac{1}{4}.$